Section 4.4 Suppose X and Y are two random variables such that (1) (2) (3) Y | X = x has a normal distribution for any real number x. E(Y | X = x) is a linear function of x, Var(Y | X = x) is a constant for each x. Then we know that (1) (2) E(Y | X = x) = Y Y + — (x – X ) X (because when a conditional mean is a linear function, that linear function must be the least squares line). Var(Y | X = x) = – 2 Y|x f 1 (x) dx =
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Section 4.4 Suppose X and Y are two random variables such that (1) (2) (3)
Section 4.4 Suppose X and Y are two random variables such that (1) (2) (3). E( Y | X = x ) is a linear function of x ,. Var( Y | X = x ) is a constant for each x. Y | X = x has a normal distribution for any real number x. Then we know that (1) (2). - PowerPoint PPT Presentation
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Section 4.4
Suppose X and Y are two random variables such that
(1)
(2)
(3) Y | X = x has a normal distribution for any real number x.
E(Y | X = x) is a linear function of x,
Var(Y | X = x) is a constant for each x.
Then we know that
(1)
(2)
E(Y | X = x) =Y
Y + — (x – X)X
(because when a conditional mean is a linear function, that linear function must be the least squares line).
Suppose X and Y are two random variables such that
(1)
(2)
(3)
Then we know that
(1)
(2)
E(Y | X = x) =Y
Y + — (x – X)X
Var(Y | X = x) = 2Y (1 – 2)
h(y | x) =
Y | X = x has a normal distribution for any real number x,
E(Y | X = x) is a linear function of x,
Var(Y | X = x) is a constant for each x.
exp
Y
– y – Y + — (x – X)X
2
22Y (1 – 2)
(2)1/2Y (1 – 2)1/2
– < y <
(3) For all x,
Now, suppose that X has a normal distribution. Then, the joint p.d.f. of (X,Y) is f(x,y) = f1(x) h(y | x) =
exp– (x – X)2
22X
(2)1/2X
exp
Y
– y – Y + — (x – X)X
2
22Y (1 – 2)
(2)1/2Y (1 – 2)1/2
exp
Y 2Y
(y – Y)2 – 2 — (x – X)(y – Y) + 2 — (x – X)2
X 2X
22Y (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2
=
=
exp
Y 2Y
(y – Y)2 – 2 — (x – X)(y – Y) + 2 — (x – X)2
X 2X
22Y (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2=
exp
y – Y x – X y – Y x – X ——— – 2 ——— ——— + 2 ——— Y X Y X
2 (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2=
2 2
exp
y – Y x – X y – Y x – X ——— – 2 ——— ——— + 2 ——— Y X Y X
2 (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2=
2 2
exp
2X Y (1 – 2)1/2
2 (1 – 2)
2 x – X——— – X
x – X y – Y y – Y 2 ——— ——— + ——— X Y Y
2
This is called a bivariate normal p.d.f., and (X, Y) are said to have a bivariate normal distribution with correlation coefficient .
(Note: Completing this derivation is Text Exercise 4.4-2.)
– < x <
– < y <
exp
2X Y (1 – 2)1/2
2 (1 – 2)
2 x – X——— – X
x – X y – Y y – Y 2 ——— ——— + ——— X Y Y
2
– < x <
– < y <
From our derivation of the bivariate normal p.d.f., we know that
(1) X has a N( X , 2X ) distribution,
(2) Y | X = x has a N( , ) distribution.Y
Y + — (x – X)X
2Y (1 – 2)
From the symmetry of the bivariate normal p.d.f., we find that
(1) Y has a distribution,
(2) X | Y = y has a N( , ) distribution.X
X + — (y – Y)Y
2X (1 – 2)
N( Y , 2Y )
Important Theorem in the Text:
If (X, Y) have a bivariate normal distribution with correlation coefficient , then X and Y are independent if and only if = 0.
Theorem 4.4-1
1.
(a)
Random variables X and Y are respectively the height (inches) and weight (lbs.) of adult males in a particular population. It is known that X and Y have a bivariate normal distribution with